Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		int n;
		Scanner s  = new Scanner(System.in);
		n = s.nextInt();
		double [] probabilities = new double[n];
		for(int i=0;i<n;i++) {
			probabilities[i] = s.nextDouble();
		}
		double [][] dp = new double[n+1][n+1];
		//dp[i][j] = probability of getting j heads from the first i coins
		dp[0][0] = 1;
		//probability of getting 0 heads from 0 coins is 0
		for(int i=1;i<dp.length;i++) {
			for(int j=0;j<=i;j++) {
				if(j==0) {
					dp[i][j]=dp[i-1][j]*(1-probabilities[i-1]);
				}
				else {
					dp[i][j] = dp[i-1][j]*(1-probabilities[i-1]) + dp[i-1][j-1] * probabilities[i-1];
				}
			}
		}
		double sum = 0;
		for(int i=n/2 + 1;i<n+1;i++) {
			sum+=dp[n][i];
		}
		System.out.println(sum);

	}

}